Let $V$ be a simple solid region oriented with outward normals that has a piecewise-smooth boundary surface $S$. $ \oiint_S \left[ y^2x \hat{\imath} + \cos(xy) \hat{\jmath} + (e^xe^z - \sin(z)) \hat{k} \right] \cdot dS$ Use the divergence theorem to rewrite the surface integral as a triple integral. $ \iiint_V $ $ \, dV$
Answer: Assume we have a simple solid region $V$ oriented with outward normals, and it has a piecewise-smooth, closed boundary surface $S$. If $F$ is a continuously differentiable vector field in $\mathbb{R}^3$, then the divergence theorem says: $ \oiint_S F \cdot dS = \iiint_V \text{div}(F) \, dV$ The given surface, boundary, and vector field satisfy the conditions for the divergence theorem. We're converting a surface integral into a triple integral, so we know $F$ and we want to find $\text{div}(F)$. $\begin{aligned} F(x, y, z) &= y^2x \hat{\imath} + \cos(xy) \hat{\jmath} + (e^xe^z - \sin(z)) \hat{k} \\ \\ \text{div}(F) &= \dfrac{\partial}{\partial x} \left[ y^2x \right] \\ \\ &+ \dfrac{\partial}{\partial y} \left[ \cos(xy) \right] \\ \\ &+ \dfrac{\partial}{\partial z} \left[ e^xe^z - \sin(z) \right] \\ \\ &= y^2 - x\sin(xy) + e^xe^z - \cos(z) \end{aligned}$ Therefore, the equivalent triple integral is: $ \iiint_V y^2 - x\sin(xy) + e^xe^z - \cos(z) \, dV$